maclaurin e^x*sin(2x)

Lösung

maclaurin

e^{x} \cdot sin (2 x)

2 x + 2 x^{2} - \frac{1}{3} x^{3} - x^{4} - \frac{19}{60} x^{5} + \dots

Schritte zur Lösung

Apply the Maclaurin Formula

= 0 + \frac{\frac{d}{d x} ( e ^{x} sin ( 2 x ) ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( e ^{x} sin ( 2 x ) ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( e ^{x} sin ( 2 x ) ) ( 0 )}{3 !} x^{3} + \dots

Evaluate Derivatives

= 0 + \frac{2}{1 !} x + \frac{4}{2 !} x^{2} + \frac{- 2}{3 !} x^{3} + \frac{- 24}{4 !} x^{4} + \frac{- 38}{5 !} x^{5} + \dots

Fasse zusammen

= 2 x + 2 x^{2} - \frac{1}{3} x^{3} - x^{4} - \frac{19}{60} x^{5} + \dots

l a pl a ce t r an s f or m f (t) = t^{2} + 6 t - 3

t an g e n t f (x) = 2 x^{4}, (- 2, 32)

\frac{d}{d x} (18 x e^{x^{2}})

\int 3 e^{x} - sin (x) d x

n \to \infty lim (\frac{( cos ( \frac{1}{n} ) - 1 )}{\frac{1}{n}})